5.1 Finding the equation of a line

If we’re given the slope, $m$ , and any point $(x{}_1,y_1)$ on the line, we can substitute this information into the formula for slope.
Let $(x{}_1,y_1)$ be the known point on the line and let $(x,y)$ be any other point on the line. Then

$\begin{array}{llll}\hfill m& =\hfill & \frac{y-y_1}{x-x_1}\hfill & \text{Multiply\hspace{0.17em}}\text{both\hspace{0.17em}}\text{sides\hspace{0.17em}}\text{by}x-x_1.\hfill \\ m(x-x_1)\hfill & =\hfill & \cancel{(x-x_1)}·\frac{y-y_1}{\cancel{x-x_1}}\hfill & \hfill \\ m(x-x_1)\hfill & =\hfill & y-y_1\hfill & \text{For\hspace{0.17em}}\text{convenience,\hspace{0.17em}}\text{we'll\hspace{0.17em}}\text{rewrite\hspace{0.17em}}\text{the\hspace{0.17em}}\text{equation}.\hfill \\ \hfill y-y_1& =\hfill & m(x-x_1)\hfill & \hfill \end{array}$

Since this equation was derived using a point and the slope of a line, it is called the point-slope form of a line.

If we are given the slope, $m$ , $\text{y-intercept},\hspace{0.17em}(0,b)$ , we can substitute this information into the formula for slope.
Let $(0,b)$ be the $\text{y-intercept\hspace{0.17em}}\text{and\hspace{0.17em}}(x,y)$ be any other point on the line. Then,

$\begin{array}{llll}\hfill m& =\hfill & \frac{y-b}{x-0}\hfill & \hfill \\ \hfill m& =\hfill & \frac{y-b}{x}\hfill & \text{Multiply\hspace{0.17em}}\text{both\hspace{0.17em}}\text{sides\hspace{0.17em}}\text{by\hspace{0.17em}}x\text{.}\hfill \\ \hfill m·x& =\hfill & \cancel{x}·\frac{y-b}{\cancel{x}}\hfill & \hfill \\ \hfill mx& =\hfill & y-b\hfill & \text{Solve\hspace{0.17em}}\text{for\hspace{0.17em}}y.\hfill \\ mx+b\hfill & =\hfill & \hfill y& \text{For\hspace{0.17em}}\text{convenience,\hspace{0.17em}\hspace{0.17em}}\text{we'll\hspace{0.17em}}\text{rewrite\hspace{0.17em}}\text{this\hspace{0.17em}}\text{equation}\text{.}\hfill \\ \hfill y& =\hfill & mx+b\hfill & \hfill \end{array}$

Since this equation was derived using the slope and the intercept, it was called the slope-intercept form of a line.